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PS: I edited the title again and I think, it's better now... :) Actually, almost the whole solving way is wrong but, I understood why... :D :)
You can check that or add some more useful links about my method, unfortunately, which I applied it wrong... Thank you. :)

Here is a problem which I've encountered and found the answer luckily, I think. But, actually, I need some better or faster ways to solve that:

The problem: $ \; \large{ 3xy+y-6x-2=0 \; , \; y=\,? } \; $

$$ My \; method: $$

$$ \large{ y(3x+1)-6x-2=0 } \\ $$ $$ \large{ X_{1}+X_{2}=\frac{-b}{a} \; \land \; X_{1}-X_{2}=\frac{ \sqrt{\Delta} } {|a|} \; \land \; \Delta=(b)^{2}-((4) \times (a) \times (c)) } $$ $$ \large{ X_{1}+X_{2}=\frac{-(-6)}{(3x+1)} \; , \; \; X_{1}-X_{2}=\frac{ \sqrt{44+24x} } {|3x+1|} \; , \; \; \Delta=44+24x } $$ $$ \large{ (X_{1}+X_{2})+(X_{1}-X_{2})=2X_{1} \\ 2X_{1}=\frac{-(-6)}{(3x+1)}+\frac{ \sqrt{44+24x} } {|3x+1|} \\ \frac{ \sqrt{44+24x}+6 } {|3x+1|}=2X_{1} \\ 8X_{1}=\sqrt{44+24x}+6 \\ (8X_{1})^{2}=(\sqrt{44})^{2}+(\sqrt{24})^{2}+(6)^{2} \\ 64X_{1}=44+24x+36 \\ 40X_{1}=80 \\ X_{1}=2 } $$ $$ \text{ Then, I recalled the first equation with } \, ''X_{1}'' \, \text{ and rewrote it down: } $$ $$ \large{ 3 \times (2) \times y + y - 6 \times (2) = 0 \\ 7y-14=0 \\ y-2=0 \\ y=2 } $$

My other question is, if you would encounter this problem in a test, which method you would try to do it as fast as you could?

Thank you very much!...

Kerim Atasoy
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    Note $-6x-2=-2(3x+1)$. So, after factoring $(3x+1)(y-2)=0$. – David Mitra Apr 26 '12 at 21:26
  • You might consider renaming the question.... – John Gowers Apr 26 '12 at 21:28
  • @Donkey_2009: I did my best for the title, please, if you have some other idea for that, make it more clear or better, thank you... :) – Kerim Atasoy Apr 26 '12 at 21:33
  • @DavidMitra: Yes, David, I see it now... Thank you very much. :) – Kerim Atasoy Apr 26 '12 at 21:34
  • Are you trying to apply the formula for the solutions of a quadratic to a linear equation? – Jyrki Lahtonen Apr 27 '12 at 13:17
  • @JyrkiLahtonen: Not exactly, sir. But, I have more idea now... :) I was not good before at Mathematics, I mean, as much as now, because, generally, I had a very poor basic knowledge and skill about Maths that comes and gets more complicated since my childhood... On the other hand, especially recently, I'm having some very good practicing opportunities such as by using this site, studying Maths for my undergraduate mathematics lessons, my eager which motivates me about Maths generally... :) I'm really having so much fun todays!... :) By the way, thank you Everyone, who helps me out a lot. :) – Kerim Atasoy Apr 27 '12 at 14:45
  • @Kerim, +1 for showing your work! If only everybody who asks here did the same ;-) – Jyrki Lahtonen Apr 27 '12 at 16:24

3 Answers3

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I would isolate $y$ on one side of the equation (that is, solve for $y$ in terms of $x$; nothing at the outset tells you $y$ has a particular value).

$$\tag{1} y(3x+1)-6x-2 \iff y(3x+1) =6x+2. $$ The next step would be to divide both sides by $3x+1$. But we can't do that if $x=-1/3$. However, if $x=-1/3$, then equation $(1)$ holds, and $y$ can be anything.

If $x\ne-1/3$, then equation $(1)$ is equivalent to $$ y={6x+2\over 3x+1}\iff y={2(3x+1)\over 3x+1}\iff y=2. $$

David Mitra
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Frankly, I have no idea what you have done above.

How about the following: Factor your equation once more to get $(3x+1)(y-2)=0$.

At least one term must be zero. So the solutions are $x=-\frac{1}{3}$ with arbitrary $y$, or $y=2$ with arbitrary $x$.

copper.hat
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  • :D So, I did it all wrong except just for the answer... :) I knew that I was just so lucky... :D :) At least, I got a good lesson by trying and sharing it with you... :) Thank you. :) – Kerim Atasoy Apr 26 '12 at 21:42
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The expression factors as $(3x+1)(y-2)$.

Alternately, we have $$y(3x+1)=6x+2.$$ Divide, which is fine unless $3x+1=0$. We get $$y=\frac{6x+2}{3x+1}=2.$$

If $3x+1=0$, then any $y$ will do. So the solutions of the given equation are (i) all pairs $(x,y)$ such that $y=2$ and (ii) all pairs $(x,y)$ such that $x=-1/3$.

André Nicolas
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